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Loss of rotational energy to thermal energy 
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#1
Nov406, 11:59 AM

P: 2

I've got a turntable whose bearing is frictionless rotating at a certain speed. A mass of clay is dropped on to it and sticks a certain distance from the center. I've been able to calculate the new angular speed of the turntable after the lump falls on it, but the follow up question is to calculate the loss of energy to thermal energy.
The bearing is frictionless, so my hunch is it has to do with the torque on the system. I've tried using the formulas for rotational kinetic energy and factoring in loss by adding a tangential component or a radial component to the right hand side to find out what the lost energy is but I'm stuck. Am I thinking from the wrong angle? (hehe) Thanks for any hints in the right direction. 


#2
Nov406, 01:34 PM

Sci Advisor
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P: 3,033

Unless you are given information about the lump of clay falling some distance or with some velocity, you may assume it had no kinetic energy before sticking. Using the angular velocities you can calculate the total kinetic energy before and after the clay sticks to the turntable. If the kinetic energy changes, the difference had to be converted into some other form of energy, and that would be thermal energy in this case. You don't need to know anything about the forces, which is a good thing because you have no way of finding them.



#3
Nov406, 01:36 PM

HW Helper
P: 1,449

In collisions we often find that some of the initial energy of the objects gets converted into other forms of energy, unless the interaction is completely elastic, which is not the case here. Since the two objects stick together after the collision we classify the collision as completely inelastic. In this case a lot of energy will be converted to heat by the deforming lump of clay.



#4
Nov406, 03:01 PM

P: 2

Loss of rotational energy to thermal energy
Awesome, that points me in the right direction.



#5
Oct1009, 06:34 PM

P: 671

The difference between the two is what's been converted to heat. For an object rotating about its center of mass, the rotational kinetic energy is given by: [tex]E_k=\tfrac{1}{2}I\omega ^2[/tex] EDIT: OOF, didn't notice I was necroposting, my apologies. 


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